In a randomized controlled trial, insecticide-treated bednets were tested as a w
ID: 3364442 • Letter: I
Question
In a randomized controlled trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 330 infants using bednets, 14 developed malaria. Among 266 infants not using bednets, 30 developed malaria. Use a 0.05 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be effective? Conduct the hypothesis test by using the results from the given display.
Differenceequalsp(1)minus
p(2)
Estimate for difference: negative 0.0703577
95
%
upper bound for difference: negative 0.02656324 Test for
differenceequals
0
(vs
less than
0):
Zequals
negative 3.27
P-Valueequals
0.001
Explanation / Answer
Given that,
sample one, x1 =14, n1 =330, p1= x1/n1=0.042
sample two, x2 =30, n2 =266, p2= x2/n2=0.113
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.042-0.113)/sqrt((0.074*0.926(1/330+1/266))
zo =-3.265
| zo | =3.265
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =3.265 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.2654 ) = 0.0011
hence value of p0.05 > 0.0011,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -3.265
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0011
TRADITIONAL METHOD
given that,
sample one, x1 =14, n1 =330, p1= x1/n1=0.0424
sample two, x2 =30, n2 =266, p2= x2/n2=0.1128
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.0424*0.9576/330) +(0.1128 * 0.8872/266))
=0.0223
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0223
=0.0438
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.0424-0.1128) ±0.0438]
= [ -0.1142 , -0.0266]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =14, n1 =330, p1= x1/n1=0.0424
sample two, x2 =30, n2 =266, p2= x2/n2=0.1128
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.0424-0.1128) ± 1.96 * 0.0223]
= [ -0.1142 , -0.0266 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ -0.1142 , -0.0266] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean P1-P2
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